# Question 6393a

May 20, 2015

You'd need 67 g of potassium nitrate to make that respective solution.

Molarity is defined as moles of solute, in your case potassium nitrate, per liters of solution.

Since you know the volume of the final solution, you can use its molarity to determine how many moles of potassium nitrate it must contain

$C = \frac{n}{V} \implies n = C \cdot V$

${n}_{K N {O}_{3}} = 1.1 \text{moles"/cancel("L") * 600 * 10^(-3)cancel("L") = "0.66 moles}$ $K N {O}_{3}$

Now use potassium nitrate's molar mass to determine how many grams would contain that many moles

0.66cancel("moles"KNO_3) * "101.1 g"/(1cancel("mole")KNO_3) = "66.7 g"#

Rounded to two sig figs, the number of sig figs you gave for the molarity of the solution, the answer is

${m}_{K N {O}_{3}} = \textcolor{g r e e n}{\text{67 g}}$